YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP_2010/86745.srs

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 QTRSRRRProof (⇔, 139 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 50 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

0(1(2(1(x)))) → 3(3(2(x)))
1(2(3(2(x)))) → 3(4(4(2(x))))
0(5(1(4(1(x))))) → 4(1(4(3(x))))
4(3(0(2(2(x))))) → 4(1(1(4(5(x)))))
5(5(1(5(2(x))))) → 4(3(5(2(x))))
0(4(3(3(4(4(1(x))))))) → 4(4(2(5(0(2(2(x)))))))
2(0(4(1(2(2(1(3(x)))))))) → 2(3(0(4(5(5(1(1(x))))))))
1(2(2(1(5(2(1(2(1(x))))))))) → 1(2(2(2(0(2(4(4(3(x)))))))))
4(5(1(4(3(4(3(5(4(3(x)))))))))) → 4(3(2(0(2(4(3(2(3(x)))))))))
1(4(1(2(5(3(4(3(3(2(2(x))))))))))) → 0(0(1(4(0(4(5(2(3(0(4(x)))))))))))
4(5(1(3(2(2(5(4(3(5(4(x))))))))))) → 4(0(1(1(5(3(5(4(2(2(4(x)))))))))))
5(1(4(0(1(5(5(3(3(0(3(2(x)))))))))))) → 3(0(5(3(2(0(1(0(4(1(2(x)))))))))))
0(4(1(1(3(3(2(5(4(2(2(1(3(x))))))))))))) → 1(0(1(4(3(4(4(2(3(4(2(2(1(x)))))))))))))
5(1(3(3(5(3(1(3(2(1(2(0(4(x))))))))))))) → 2(2(3(2(3(2(3(5(2(5(1(4(x))))))))))))
0(5(3(5(3(3(3(3(4(5(5(5(4(4(x)))))))))))))) → 0(1(0(3(2(3(4(0(5(5(2(4(0(x)))))))))))))
1(5(0(1(0(4(4(2(2(3(4(1(4(1(x)))))))))))))) → 3(4(5(0(0(4(0(3(5(0(4(1(5(4(x))))))))))))))
5(1(0(5(2(2(2(3(3(2(5(1(5(1(x)))))))))))))) → 5(4(0(1(4(3(2(2(3(3(3(5(5(1(x))))))))))))))
5(1(2(2(4(0(2(4(2(5(2(1(4(0(5(x))))))))))))))) → 0(1(4(3(0(5(3(4(3(3(1(4(1(5(x))))))))))))))
0(4(0(3(2(0(2(1(2(0(0(2(4(2(3(4(x)))))))))))))))) → 3(2(1(3(3(4(5(5(4(0(3(2(1(2(3(x)))))))))))))))
1(5(1(3(3(3(0(4(0(2(3(1(5(1(4(2(x)))))))))))))))) → 3(4(0(5(0(4(4(0(2(1(3(1(4(0(4(2(x))))))))))))))))
5(1(1(0(0(3(2(5(0(3(4(2(1(2(5(1(x)))))))))))))))) → 5(1(3(2(1(0(1(0(5(5(3(1(1(4(1(0(x))))))))))))))))
3(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x))))))))))))))))) → 3(2(3(1(0(2(0(2(4(5(5(4(1(0(0(2(5(0(x))))))))))))))))))
4(1(0(3(0(4(3(2(2(1(3(2(2(4(0(2(4(x))))))))))))))))) → 4(2(5(3(3(3(1(2(4(5(3(5(3(5(1(3(4(x)))))))))))))))))
3(0(0(2(1(1(3(5(1(2(2(2(5(1(0(0(0(1(x)))))))))))))))))) → 3(5(1(0(4(0(1(2(2(5(0(3(4(3(5(5(4(3(x))))))))))))))))))
4(5(5(1(5(3(5(3(2(0(4(4(2(1(0(3(5(3(x)))))))))))))))))) → 4(4(0(5(1(3(5(5(3(4(4(0(0(4(3(0(0(0(x))))))))))))))))))
4(5(5(2(5(1(0(2(1(0(1(4(4(4(2(1(5(1(x)))))))))))))))))) → 4(3(2(3(0(2(5(3(4(1(4(4(5(1(1(4(0(x)))))))))))))))))
0(2(0(2(2(0(1(1(2(4(1(1(0(3(3(2(1(4(1(4(x)))))))))))))))))))) → 2(2(3(3(0(2(1(3(5(3(4(4(1(2(4(4(4(4(4(0(x))))))))))))))))))))
3(4(1(1(0(3(4(0(5(5(5(5(3(5(2(3(2(3(1(3(x)))))))))))))))))))) → 3(4(3(4(3(1(0(1(1(4(5(5(2(3(2(3(0(2(3(x)))))))))))))))))))
4(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 4(4(3(1(4(4(0(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x)))))))))))))))))))))
5(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x))))))))))))))))))))) → 5(0(3(2(2(0(4(1(1(5(3(0(1(5(0(1(3(2(2(3(x))))))))))))))))))))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0(x1)) = 61 + x1   
POL(1(x1)) = 70 + x1   
POL(2(x1)) = 76 + x1   
POL(3(x1)) = 77 + x1   
POL(4(x1)) = 64 + x1   
POL(5(x1)) = 60 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

0(1(2(1(x)))) → 3(3(2(x)))
1(2(3(2(x)))) → 3(4(4(2(x))))
0(5(1(4(1(x))))) → 4(1(4(3(x))))
4(3(0(2(2(x))))) → 4(1(1(4(5(x)))))
5(5(1(5(2(x))))) → 4(3(5(2(x))))
2(0(4(1(2(2(1(3(x)))))))) → 2(3(0(4(5(5(1(1(x))))))))
1(2(2(1(5(2(1(2(1(x))))))))) → 1(2(2(2(0(2(4(4(3(x)))))))))
4(5(1(4(3(4(3(5(4(3(x)))))))))) → 4(3(2(0(2(4(3(2(3(x)))))))))
1(4(1(2(5(3(4(3(3(2(2(x))))))))))) → 0(0(1(4(0(4(5(2(3(0(4(x)))))))))))
4(5(1(3(2(2(5(4(3(5(4(x))))))))))) → 4(0(1(1(5(3(5(4(2(2(4(x)))))))))))
5(1(4(0(1(5(5(3(3(0(3(2(x)))))))))))) → 3(0(5(3(2(0(1(0(4(1(2(x)))))))))))
0(4(1(1(3(3(2(5(4(2(2(1(3(x))))))))))))) → 1(0(1(4(3(4(4(2(3(4(2(2(1(x)))))))))))))
5(1(3(3(5(3(1(3(2(1(2(0(4(x))))))))))))) → 2(2(3(2(3(2(3(5(2(5(1(4(x))))))))))))
0(5(3(5(3(3(3(3(4(5(5(5(4(4(x)))))))))))))) → 0(1(0(3(2(3(4(0(5(5(2(4(0(x)))))))))))))
1(5(0(1(0(4(4(2(2(3(4(1(4(1(x)))))))))))))) → 3(4(5(0(0(4(0(3(5(0(4(1(5(4(x))))))))))))))
5(1(2(2(4(0(2(4(2(5(2(1(4(0(5(x))))))))))))))) → 0(1(4(3(0(5(3(4(3(3(1(4(1(5(x))))))))))))))
0(4(0(3(2(0(2(1(2(0(0(2(4(2(3(4(x)))))))))))))))) → 3(2(1(3(3(4(5(5(4(0(3(2(1(2(3(x)))))))))))))))
1(5(1(3(3(3(0(4(0(2(3(1(5(1(4(2(x)))))))))))))))) → 3(4(0(5(0(4(4(0(2(1(3(1(4(0(4(2(x))))))))))))))))
5(1(1(0(0(3(2(5(0(3(4(2(1(2(5(1(x)))))))))))))))) → 5(1(3(2(1(0(1(0(5(5(3(1(1(4(1(0(x))))))))))))))))
4(1(0(3(0(4(3(2(2(1(3(2(2(4(0(2(4(x))))))))))))))))) → 4(2(5(3(3(3(1(2(4(5(3(5(3(5(1(3(4(x)))))))))))))))))
3(0(0(2(1(1(3(5(1(2(2(2(5(1(0(0(0(1(x)))))))))))))))))) → 3(5(1(0(4(0(1(2(2(5(0(3(4(3(5(5(4(3(x))))))))))))))))))
4(5(5(1(5(3(5(3(2(0(4(4(2(1(0(3(5(3(x)))))))))))))))))) → 4(4(0(5(1(3(5(5(3(4(4(0(0(4(3(0(0(0(x))))))))))))))))))
4(5(5(2(5(1(0(2(1(0(1(4(4(4(2(1(5(1(x)))))))))))))))))) → 4(3(2(3(0(2(5(3(4(1(4(4(5(1(1(4(0(x)))))))))))))))))
0(2(0(2(2(0(1(1(2(4(1(1(0(3(3(2(1(4(1(4(x)))))))))))))))))))) → 2(2(3(3(0(2(1(3(5(3(4(4(1(2(4(4(4(4(4(0(x))))))))))))))))))))
3(4(1(1(0(3(4(0(5(5(5(5(3(5(2(3(2(3(1(3(x)))))))))))))))))))) → 3(4(3(4(3(1(0(1(1(4(5(5(2(3(2(3(0(2(3(x)))))))))))))))))))


(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

0(4(3(3(4(4(1(x))))))) → 4(4(2(5(0(2(2(x)))))))
5(1(0(5(2(2(2(3(3(2(5(1(5(1(x)))))))))))))) → 5(4(0(1(4(3(2(2(3(3(3(5(5(1(x))))))))))))))
3(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x))))))))))))))))) → 3(2(3(1(0(2(0(2(4(5(5(4(1(0(0(2(5(0(x))))))))))))))))))
4(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 4(4(3(1(4(4(0(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x)))))))))))))))))))))
5(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x))))))))))))))))))))) → 5(0(3(2(2(0(4(1(1(5(3(0(1(5(0(1(3(2(2(3(x))))))))))))))))))))

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

01(4(3(3(4(4(1(x))))))) → 41(4(2(5(0(2(2(x)))))))
01(4(3(3(4(4(1(x))))))) → 41(2(5(0(2(2(x))))))
01(4(3(3(4(4(1(x))))))) → 51(0(2(2(x))))
01(4(3(3(4(4(1(x))))))) → 01(2(2(x)))
51(1(0(5(2(2(2(3(3(2(5(1(5(1(x)))))))))))))) → 51(4(0(1(4(3(2(2(3(3(3(5(5(1(x))))))))))))))
51(1(0(5(2(2(2(3(3(2(5(1(5(1(x)))))))))))))) → 41(0(1(4(3(2(2(3(3(3(5(5(1(x)))))))))))))
51(1(0(5(2(2(2(3(3(2(5(1(5(1(x)))))))))))))) → 01(1(4(3(2(2(3(3(3(5(5(1(x))))))))))))
51(1(0(5(2(2(2(3(3(2(5(1(5(1(x)))))))))))))) → 41(3(2(2(3(3(3(5(5(1(x))))))))))
51(1(0(5(2(2(2(3(3(2(5(1(5(1(x)))))))))))))) → 31(2(2(3(3(3(5(5(1(x)))))))))
51(1(0(5(2(2(2(3(3(2(5(1(5(1(x)))))))))))))) → 31(3(3(5(5(1(x))))))
51(1(0(5(2(2(2(3(3(2(5(1(5(1(x)))))))))))))) → 31(3(5(5(1(x)))))
51(1(0(5(2(2(2(3(3(2(5(1(5(1(x)))))))))))))) → 31(5(5(1(x))))
51(1(0(5(2(2(2(3(3(2(5(1(5(1(x)))))))))))))) → 51(5(1(x)))
31(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x))))))))))))))))) → 31(2(3(1(0(2(0(2(4(5(5(4(1(0(0(2(5(0(x))))))))))))))))))
31(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x))))))))))))))))) → 31(1(0(2(0(2(4(5(5(4(1(0(0(2(5(0(x))))))))))))))))
31(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x))))))))))))))))) → 01(2(0(2(4(5(5(4(1(0(0(2(5(0(x))))))))))))))
31(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x))))))))))))))))) → 01(2(4(5(5(4(1(0(0(2(5(0(x))))))))))))
31(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x))))))))))))))))) → 41(5(5(4(1(0(0(2(5(0(x))))))))))
31(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x))))))))))))))))) → 51(5(4(1(0(0(2(5(0(x)))))))))
31(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x))))))))))))))))) → 51(4(1(0(0(2(5(0(x))))))))
31(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x))))))))))))))))) → 41(1(0(0(2(5(0(x)))))))
31(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x))))))))))))))))) → 01(0(2(5(0(x)))))
31(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x))))))))))))))))) → 01(2(5(0(x))))
31(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x))))))))))))))))) → 51(0(x))
31(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x))))))))))))))))) → 01(x)
41(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 41(4(3(1(4(4(0(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x)))))))))))))))))))))
41(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 41(3(1(4(4(0(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x))))))))))))))))))))
41(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 31(1(4(4(0(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x)))))))))))))))))))
41(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 41(4(0(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x)))))))))))))))))
41(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 41(0(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x))))))))))))))))
41(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 01(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x)))))))))))))))
41(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 51(3(5(2(1(4(2(4(1(0(2(4(5(0(x))))))))))))))
41(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 31(5(2(1(4(2(4(1(0(2(4(5(0(x)))))))))))))
41(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 51(2(1(4(2(4(1(0(2(4(5(0(x))))))))))))
41(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 41(2(4(1(0(2(4(5(0(x)))))))))
41(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 41(1(0(2(4(5(0(x)))))))
41(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 01(2(4(5(0(x)))))
41(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 41(5(0(x)))
41(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 51(0(x))
51(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x))))))))))))))))))))) → 51(0(3(2(2(0(4(1(1(5(3(0(1(5(0(1(3(2(2(3(x))))))))))))))))))))
51(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x))))))))))))))))))))) → 01(3(2(2(0(4(1(1(5(3(0(1(5(0(1(3(2(2(3(x)))))))))))))))))))
51(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x))))))))))))))))))))) → 31(2(2(0(4(1(1(5(3(0(1(5(0(1(3(2(2(3(x))))))))))))))))))
51(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x))))))))))))))))))))) → 01(4(1(1(5(3(0(1(5(0(1(3(2(2(3(x)))))))))))))))
51(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x))))))))))))))))))))) → 41(1(1(5(3(0(1(5(0(1(3(2(2(3(x))))))))))))))
51(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x))))))))))))))))))))) → 51(3(0(1(5(0(1(3(2(2(3(x)))))))))))
51(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x))))))))))))))))))))) → 31(0(1(5(0(1(3(2(2(3(x))))))))))
51(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x))))))))))))))))))))) → 01(1(5(0(1(3(2(2(3(x)))))))))
51(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x))))))))))))))))))))) → 51(0(1(3(2(2(3(x)))))))
51(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x))))))))))))))))))))) → 01(1(3(2(2(3(x))))))
51(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x))))))))))))))))))))) → 31(2(2(3(x))))
51(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x))))))))))))))))))))) → 31(x)

The TRS R consists of the following rules:

0(4(3(3(4(4(1(x))))))) → 4(4(2(5(0(2(2(x)))))))
5(1(0(5(2(2(2(3(3(2(5(1(5(1(x)))))))))))))) → 5(4(0(1(4(3(2(2(3(3(3(5(5(1(x))))))))))))))
3(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x))))))))))))))))) → 3(2(3(1(0(2(0(2(4(5(5(4(1(0(0(2(5(0(x))))))))))))))))))
4(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x))))))))))))))))))))) → 4(4(3(1(4(4(0(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x)))))))))))))))))))))
5(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x))))))))))))))))))))) → 5(0(3(2(2(0(4(1(1(5(3(0(1(5(0(1(3(2(2(3(x))))))))))))))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 51 less nodes.

(6) TRUE